Voltage Drop is the loss of voltage in a non-ideal conductor. In other words, conductors have an impedance, a resistance and a reactance, that causes the voltage magnitude at the supply end of a conductor to be different than at the load end of a conductor when there is current flowing. Voltage drop can become a serious design issue. If we were to lose 50% of the voltage in a cable before it reaches the load, we can't reasonably expect a load to operate correctly.
Voltage Drop in a Nutshell
The National Electrical Code addresses voltage drop only through informational notes (sections that aren't enforceable). It says that branch circuits should have a voltage drop less than 3%, and that total circuits (feeders + branch circuits) should have a voltage drop less than 5% to generally be okay. The reason this information only exists in information notes is twofold. First, the NEC isn't a guide for good designs, only a guide to keep designs safe. If voltage drop is too high, your installation probably won't become unsafe; it just won't work the way you want it to. Second, the NEC requires electrical equipment to be installed in accordance with its instructions. If equipment has its safety impacted by a voltage range and this is required in the instructions, then this is also required by Code (See Article 110).
Voltage drop calculations differ for DC circuits, AC single-phase circuits, and AC three-phase circuits. The relevant formulas for each are:
DC Circuits: V = 2 I L rdc
AC Single-Phase Circuits: V = 2 I L z
AC Three-Phase Circuits: V = √(3) I L z
Where:
V is the voltage drop across a conductor in Volts, as measured between line conductors
I is the load current flowing through a conductor in Amps. Modifications for continuous loading should not be included.
L is the one-way length of the circuit in feet. This length is the same if your circuit has 2 wires or 3 wires.
rdc is the DC resistance of the conductor in Ohms/foot
z is the effective AC impedance of the conductor in Ohms/foot
The effective impedance z is dependent upon the power factor of the circuit, which may be difficult to know in advance, depending on the situation. A conservative value to be used for the effective impedance is:
z = √( rac^2 + x^2)
Where:
rac is the AC resistance of the conductor in Ohms/foot
x is the reactance of the conductor in Ohms/foot
Values for conductor DC resistances and AC impedances (reactances and resistances) can be taken from NEC Chapter 9 Tables 8 and 9 for a wide range of installation methods. If available, manufacturer data is even better. Values from Chapter 9 are provided at an operating temperature of 75°C. If a different operating temperature is desired for more accurate computation of the voltage drop, the DC or AC resistance can be computed using the following equation:
r' = r (1 + a (T - 75°C) )
Where:
r' is the temperature-corrected resistance of the conductor
r is the 75°C resistance of the conductor
a is a temperature coefficient defined as .00323/°C for Copper and .00330/°C for Aluminum
T is the desired operating temperature for correction in °C
It is important to recognize the limitations of the values provided in the NEC. Chapter 9 Tables 8 and 9 are based on low voltage configurations (600V or less). Medium voltage circuits may have different values for resistance and reactance than those presented in Chapter 9 Table 9. Unlike low voltage circuits, medium voltage conductors often have metallic shields and dielectric materials that contribute to resistive losses in the conductor. Additionally, reactances can vary substantially based on the installation methods of the conductors. The values in Chapter 9 Table 9 will be approximately accurate for medium voltage conductors if installed in a configuration similar to the ones described for 600V circuits, but should not be used if a high level of precision is required.
Example: What is the approximate voltage drop across a three-phase, 480V circuit running 1000' in PVC conduit with a load current of 200A and a conductor type of 500 KCMIL Copper operating at 75°C?
Solution: Begin by identifying the correct type of circuit. Since this is a three-phase AC circuit, voltage drop is computed as:
V = √(3) I L z
Next, identify the obvious variables:
L is the one-way length of the circuit, 1000'
I is the load current, 200A
Then, move onto the less obvious variable, z. We don't know the power factor, but we can approximate the effective impedance conservatively as:
z = √( rac^2 + x^2)
rac and x can be taken from NEC Chapter 9 Table 9 for 500 KCMIL copper conductors in PVC conduit. rac = .027 Ohms/kft and x = .039 Ohms/kft. Putting everything together:
z = √( .027^2 + .039^2) = .0474 Ohms/kft
V = √(3) * 200A * 1 kft * .0474 Ohms/kft = 16.4 V
We can normzalize this voltage drop against the line-line voltage for the system, 480V, to get a sense if our voltage drop is acceptable:
16.4 / 480 = 3.4%
This voltage drop exceeds the limits set by the NEC for branch circuits but is less than the limits for the branch circuit and feeder together. Depending on the equipment requirements, this may be an issue.
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